In the reaction between 97.49 grams of zinc and 2.53 moles of HCl, what is the maximum mass in grams of ZnCl2, 136.29 g/mol, produced? Report your answer with 2 decimal places.

Question

zn(s) +2HCl(aq)-> H2(g)+ ZnCl2(aq)

Answer 1

Zn(s) + 2HCl(aq) ==> H₂(g) + ZnCl₂(aq) ... balanced equation

Since we are given the amounts of BOTH reactants, we must find which reactant is limiting. An easy way to do this is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant. Thus ...

For Zn we have 97.49 g Zn x 1 mol Zn / 65.38 g = 1.49 1 mols Zn (÷1->1.49)

For HCl we have 2.53 mols (÷2->1.26)

Since 1.26 is less than 1.49, HCl is the limiting reactant and the moles of HCl (2.53 mols) will determine the maximum amount of ZnCl₂ that can be formed.

Maximum amount of ZnCl₂ formed = 2.53 mols HCl x 1 mol ZnCl₂ / 2 mol HCl x 136.29 g / mol = 172.41 g ZnCl₂ (to 2 decimal places)

Answer 2

Do each stoichiometric calculation of the product produced given given each reactant is limiting. The smallest amount will be correct and indicate the limiting reactant:

2.53 moles HCl *(1 ZnCl₂/ 2 HCl) (136.29 g ZnCl₂/mole) = g ZnCl₂ from HCl

97.49 g Zn *(1 mole/65.38 g Zn)(1 ZnCl₂/1 Zn)(136.29 g ZnCl₂/mole) = g ZnCl₂ from Zn